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I simplified $$\sqrt{64x^4y^8z^6}$$ by taking the square root of $64$ (getting $8$), $x^4$ (getting $x^2$), $y^8$ (getting $y^4$), and $z^6$ (getting $z^3$). My answer, $8x^2y^4z^3$, not quite right and I am having a hard time understanding why not.

I even placed the problem into WolframAlpha.com (here) and got the same answer, which was really confusing. Any pointers of what I might be doing wrong?

Image of Problem

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2 Answers 2

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As per the hint, you have to be careful since $z^3$ is negative if $z<0$ and the square root refers to the principal (nonnegative) root. Wolfram also mentions the assumption that $x,y,z$ are positive in their simplification to your answer. But for arbitrary real $x,y,z$ we have

$$\sqrt{64x^4y^8z^6}=8x^2y^4|z|^3.$$

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    $\begingroup$ Ahhhh, yes. I didn't realize that one part of the answer could have an absolute value specifically for it. Thank you for that. $\endgroup$
    – יהודה
    Commented Mar 9, 2022 at 6:23
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    $\begingroup$ @יהודה no prob, feel free to accept the answer if it helps $\endgroup$ Commented Mar 9, 2022 at 6:24
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    $\begingroup$ I was trying to accept but it said I had to wait 5 mins lol $\endgroup$
    – יהודה
    Commented Mar 9, 2022 at 6:26
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It’s conventional for $\sqrt n$ to give you the positive square root of $n$, but in this case they might be expecting you to give the answer in the form $\pm {8x^2y^4z^3}$

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  • $\begingroup$ Thank you for that. Sheesh, so many different ways to answer a question. Fascinating actually. $\endgroup$
    – יהודה
    Commented Mar 9, 2022 at 6:28
  • $\begingroup$ You probably want that without the \sqrt part - it happened already. $\endgroup$
    – tevemadar
    Commented Mar 9, 2022 at 16:36
  • $\begingroup$ @tevemadar thanks $\endgroup$
    – tomi
    Commented Mar 10, 2022 at 11:38

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